Let $\mathcal A_0$ be an algebra of subsets of a non-empty set $X.$ Let $\mu$ be a finitely additive function from $\mathcal A_0$ to $[0,\infty]$ such that $\mu (\varnothing) = 0.$ Define $\mu^* : \mathcal P (X) \longrightarrow [0,\infty]$ by $$\mu^*(A) : = \inf \left \{\sum\limits_{n=1}^{\infty} \mu (A_n)\ \bigg |\ A \subseteq \bigcup\limits_{n=1}^{\infty} A_n, A_n \in \mathcal A_0 \right \}.$$ Does $\mu^* = \mu$ on $\mathcal A_0\ $?
If $\mu$ were countably-additive then I know that it's true. But for finitely additive set functions $\mu$ I don't have any idea. One direction is clear namely $\mu^*(A) \leq \mu (A).$ Is the other direction necessarily true? Any help in this regard would be greatly appreciated.
Thanks for your time.
I got a counter-example on my own. Here it is $:$
Let $X$ be a countably infinite set. Consider the collection $\mathcal A_0$ consisting of subsets of $X$ which are either finite or their complements are finite. Then it is very easy to show that $\mathcal A_0$ is an algebra of subsets of $X.$ Define a set function $\mu : \mathcal P(X) \longrightarrow [0,\infty]$ by $\mu(A) = 0,$ if $A$ is finite and $\mu(A) = 1,$ if $A^c$ is finite. Then it is easy to check that $\mu$ is a finitely additive set function with $\mu (\varnothing) = 0.$ Now take any countably infinite set $C$ whose complement is finite. Then $\mu^*(C) = 0$ but $\mu(C) = 1.$
(Since countably infinite sets can be covered by countable union of singletons each of whose size is $0$ with respect to $\mu$ by the definition of $\mu$).